We now take \(p\)  an odd prime. If \(n = 2k\)  then for the form
in (REF ) to be \(k\) -isotropic we
need \(l=k\)  and \(m=0\) . There is only one such form up to
\({\mathbb {Q}}_p\) -equivalence. It has determinant \(d(Q) = (-1)^k\)  and
Hasse-Minkowski invariant \(c(Q) = 1\) . This gives the formula for
\(\rho _p(k,2k)\)  as stated. If \(n = 2k+1\)  then for \(k\) -isotropy we
need \(l=k\)  and \(m=1\) . The anisotropic form in (REF ) is
\(f(x_n) = a x_n^2\)  for some \(a \in \lbrace 1,u,p,up\rbrace \) . This gives four
\({\mathbb {Q}}_p\) -equivalence classes of forms, with invariants
\(d(Q) = (-1)^ka\)  and \(c(Q) = (-1,a)^k\) . Finally we take \(n =2k+2\) . For \(Q\)  not to be \(k\) -isotropic we need \(l = k-1\)  and
\(m = 4\) . The rank 4 anisotropic form \(f\)  has determinant
\(d(f) = 1\)  and Hasse-Minkowski invariant \(c(f) = -1\)  (see
[1]}). It follows that \(d(Q) = (-1)^{k-1}\)  and
\(c(Q) = -1\) , giving the result as stated.
